3.47 \(\int \frac{a+b \tan ^{-1}(c x)}{x (d+i c d x)} \, dx\)

Optimal. Leaf size=54 \[ \frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{\log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

[Out]

((a + b*ArcTan[c*x])*Log[2 - 2/(1 + I*c*x)])/d + ((I/2)*b*PolyLog[2, -1 + 2/(1 + I*c*x)])/d

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Rubi [A]  time = 0.0711642, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {4868, 2447} \[ \frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{\log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(x*(d + I*c*d*x)),x]

[Out]

((a + b*ArcTan[c*x])*Log[2 - 2/(1 + I*c*x)])/d + ((I/2)*b*PolyLog[2, -1 + 2/(1 + I*c*x)])/d

Rule 4868

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[((a + b*ArcTan[c*x]
)^p*Log[2 - 2/(1 + (e*x)/d)])/d, x] - Dist[(b*c*p)/d, Int[((a + b*ArcTan[c*x])^(p - 1)*Log[2 - 2/(1 + (e*x)/d)
])/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{x (d+i c d x)} \, dx &=\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{(b c) \int \frac{\log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{i b \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0556677, size = 102, normalized size = 1.89 \[ \frac{i b \text{PolyLog}(2,-i c x)}{2 d}-\frac{i b \text{PolyLog}(2,i c x)}{2 d}+\frac{i b \text{PolyLog}\left (2,-\frac{c x+i}{-c x+i}\right )}{2 d}+\frac{\log \left (\frac{2 i}{-c x+i}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTan[c*x])/(x*(d + I*c*d*x)),x]

[Out]

(a*Log[x])/d + ((a + b*ArcTan[c*x])*Log[(2*I)/(I - c*x)])/d + ((I/2)*b*PolyLog[2, (-I)*c*x])/d - ((I/2)*b*Poly
Log[2, I*c*x])/d + ((I/2)*b*PolyLog[2, -((I + c*x)/(I - c*x))])/d

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Maple [B]  time = 0.051, size = 193, normalized size = 3.6 \begin{align*} -{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d}}-{\frac{ia\arctan \left ( cx \right ) }{d}}+{\frac{a\ln \left ( cx \right ) }{d}}-{\frac{b\arctan \left ( cx \right ) \ln \left ( cx-i \right ) }{d}}+{\frac{b\arctan \left ( cx \right ) \ln \left ( cx \right ) }{d}}+{\frac{{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) }{d}}-{\frac{{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) }{d}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( 1+icx \right ) }{d}}-{\frac{{\frac{i}{2}}b{\it dilog} \left ( 1-icx \right ) }{d}}+{\frac{{\frac{i}{2}}b\ln \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) \ln \left ( cx-i \right ) }{d}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( -{\frac{i}{2}} \left ( cx+i \right ) \right ) }{d}}-{\frac{{\frac{i}{4}}b \left ( \ln \left ( cx-i \right ) \right ) ^{2}}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/x/(d+I*c*d*x),x)

[Out]

-1/2*a/d*ln(c^2*x^2+1)-I*a/d*arctan(c*x)+a/d*ln(c*x)-b/d*arctan(c*x)*ln(c*x-I)+b/d*arctan(c*x)*ln(c*x)+1/2*I*b
/d*ln(c*x)*ln(1+I*c*x)-1/2*I*b/d*ln(c*x)*ln(1-I*c*x)+1/2*I*b/d*dilog(1+I*c*x)-1/2*I*b/d*dilog(1-I*c*x)+1/2*I*b
/d*ln(-1/2*I*(c*x+I))*ln(c*x-I)+1/2*I*b/d*dilog(-1/2*I*(c*x+I))-1/4*I*b/d*ln(c*x-I)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \, b{\left (\frac{i \, \arctan \left (c x\right )^{2}}{d} - 2 \, \int \frac{\arctan \left (c x\right )}{c^{2} d x^{3} + d x}\,{d x}\right )} - a{\left (\frac{\log \left (i \, c x + 1\right )}{d} - \frac{\log \left (x\right )}{d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x),x, algorithm="maxima")

[Out]

-1/2*b*(I*arctan(c*x)^2/d - 2*integrate(arctan(c*x)/(c^2*d*x^3 + d*x), x)) - a*(log(I*c*x + 1)/d - log(x)/d)

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Fricas [A]  time = 2.59776, size = 109, normalized size = 2.02 \begin{align*} \frac{-i \, b{\rm Li}_2\left (\frac{c x + i}{c x - i} + 1\right ) + 2 \, a \log \left (x\right ) - 2 \, a \log \left (\frac{c x - i}{c}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x),x, algorithm="fricas")

[Out]

1/2*(-I*b*dilog((c*x + I)/(c*x - I) + 1) + 2*a*log(x) - 2*a*log((c*x - I)/c))/d

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/x/(d+I*c*d*x),x)

[Out]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \left (c x\right ) + a}{{\left (i \, c d x + d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x/(d+I*c*d*x),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)/((I*c*d*x + d)*x), x)